如何避免两次走树

Question

I have the following method to initialize a tree:

function _initTree(treeObj, options){ 
    treeObj = treeObj || []; 
    if(!zpJSUtils.isArray(treeObj)) throw new Error("Input tree must be an array"); 
    this.tree = treeObj; 
    this.rootNode = treeObj[0]; 
    _buildNodes.call(this, treeObj); 
} 

The init method walks the tree to initialize all objects as nodes:

function _buildNodes(treeObj){ 
    traverseTree.call(treeObj, function(nodeObj){
        node = new Node(nodeObj); 
        return true; 
    }, this); 
} 

The traverseTree method (selected from a switch):

function _walkPreOrder(tree, callback, ctx){ 
    tree.forEach(function(node, idx){ 
        var continueWalk = callback.call(ctx, node); 
        if(continueWalk){
            if(node.hasChildren()){ 
                _walkPreOrder.call(this, node); 
            }; 
        }; 
    }, this); 
}

The question is this.
I want to calculate the height of the tree on init, which is the number of edges on the longest downward path between the root node and a leaf.
So it seems that in order to calculate the height, I'll have to walk the tree again (to find the level of the deepest nested element).

The format of the json is this:

var json = [{ text: "root", children: [
    {id: "id_1", text: "node_1", children:[
        {id: "id_c1", text: "node_c1"}, 
        {id: "id_c2", text: "node_c2", children: [
            {id: "id_c2_c1", text: "node_c2_c1"}, 
            {id: "id_c2_c2", text: "node_c2_c2"}, 
            {id: "id_c2_c3", text: "node_c2_c3"}]},   
        {id: "id_c3", text: "node_c3"}]}, 
    {id: "id_2", text: "node_2"}]}]; 

The problem is that the _walkPreOrder could accept multiple callbacks (eg array), but that would result in an dubious api (since passing in multiple functions makes the possibility to continue walking the tree when the condition is true/false dubious).

What would be a good way to solve this problem, ie to avoid multiple iterations?

Answer 1

1. On the first time you walk the tree while creating the child nodes, either return or mark each child with its max-height.
2. Then upon visiting all children, compare the maximum of all the child heights, and then mark/return the current node with 1 more than the max height of its children.